Question

Let the matrix $A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$ and the matrix $B_0=A^{49}+2A^{98}$. If $B_n=\text{Adj}\left(B_{n-1}\right)$ for all $n\ge 1$, then $\text{det}\left(B_4\right)$ is equal to :

• A. $3^{28}$
• B. $3^{30}$
• C. $3^{32}$
• D. $3^{36}$

Answer 1

Answer: (C)

$A^2=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$
$=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$
$a\leftrightarrow R_2$
$-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$
$R_2\leftrightarrow R_3$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
$B_0=A^{49}+2A^{98}$
$=A+2I$
$B_n=\text{Adj}\left(B_n-1\right)$
$B_4=\text{Adj}(\text{Adj}\left(\text{Adj}\left({\text{Adj}}_0\right)\right)$
$={\left|B_0\right|}^{(n-1{)}^4}$
$={\left|B_0\right|}^{16}$
$B_0=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]+\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$
$=\left[\begin{array}{ccc}2& 1& 0\\ 0& 2& 1\\ 1& 0& 2\end{array}\right]$
$=2(4-0)-1(0-1)$
$=9$
$B_4(9{)}^{16}=(3{)}^{32}$